Question

就是x^4/(1+x^4)的不定积分怎样算？？？

Answer 1

本题技巧高
∫x^4/(1+x^4) dx
=(1/2)∫ [(1-x²)+(1+x²)]/(1+x^4) dx
=(1/2)∫ (1-x²)/(1+x^4) dx + (1/2)∫ (1+x²)/(1+x^4) dx
分子分母同除x²
=(1/2)∫ (1/x²-1)/(x²+1/x²) dx + (1/2)∫ (1/x²+1)/(x²+1/x²) dx
=-(1/2)∫ 1/(x²+1/x²+2-2) d(x+1/x) + (1/2)∫ 1/(x²+1/x²-2+2) d(x-1/x)
=-(1/2)∫ 1/[(x+1/x)²-2] d(x+1/x) + (1/2)∫ 1/[(x-1/x)²+2] d(x-1/x)
=-(√2/8)ln|(x+1/x-√2)/(x+1/x+√2)| + (√2/4)arctan[(x-1/x)/√2] + C

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